10.2 上午 考试

Posted 2020-10-09 A_LEAF

T1

移项，然后离散，dp即可

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <ctime>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
const int N=200006;
const int N2=500006;

struct LI
{
    int pos,val,flag;
    bool friend operator < (LI a,LI b)
    {
        return a.val<b.val;
    }
}li[N2];
int con;

struct son
{
    int x,w,v[2];
    bool friend operator < (son a,son b)
    {
        return a.x<b.x;
    }
}ji[N];

int n;
int f[N];
int ans;
int now;

void lisan()
{
    for(int i=1;i<=n;++i)
    {
        li[++con]=(LI){i,ji[i].x-ji[i].w,0};
        li[++con]=(LI){i,ji[i].x+ji[i].w,1};
    }
    sort(li+1,li+1+con);
    li[0].val=-0x7fffffff;
    for(int i=1;i<=con;++i)
    {
        if(li[i].val==li[i-1].val)
            ji[li[i].pos].v[li[i].flag]=now;
        else
            ji[li[i].pos].v[li[i].flag]=++now;
    }
}

void out11()
{
    printf("\n");
    for(int i=1;i<=n;++i)
        printf("%d ",f[i]);
    printf("\n");
}

int c[N2];
inline void add(int pos,int vv)
{
    for(int i=pos;i<=now;i+=(i&(-i)) )
        if(c[i]<vv)
            c[i]=vv;
}
inline int qq(int pos)
{
    int ans=0;
    for(int i=pos;i>0;i-=(i&(-i)) )
        if(ans<c[i])
            ans=c[i];
    return ans;
}

void dp()
{
    int temp;
    for(int i=1;i<=n;++i)
    {
        temp=qq(ji[i].v[0]);
        f[i]=temp+1;
        if(ans<f[i])
            ans=f[i];
        add(ji[i].v[1],f[i]);
    }

    //out11();
}

int main(){

    //freopen("T1.in","r",stdin);
    //freopen("T1.out","w",stdout);

    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%d%d",&ji[i].x,&ji[i].w);
    sort(ji+1,ji+1+n);
    lisan();
    dp();
    cout<<ans;
}

T1

 

 

 

T2

分块

 

#pragma GCC optimize("O2")
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
const int N=100006;

int n,m;
int a[N];

int fen,dui[N],L[1006],R[1006];
int mx[1006];
ll ji[1006];

void reset(int x)
{
    mx[x]=ji[x]=0;
    for(int i=L[x];i<=R[x];++i)
    {
        if(mx[x]<a[i])
            mx[x]=a[i];
        ji[x]+=a[i];
    }
}

void mm(int l,int r,int mod)
{
    int q1=min(R[dui[l]],r);
    for(int i=l;i<=q1;++i)
        a[i]%=mod;
    reset(dui[l]);
    if(dui[l]!=dui[r])
    {
        for(int i=L[dui[r]];i<=r;++i)
            a[i]%=mod;
        reset(dui[r]);
    }
    for(int i=dui[l]+1;i<dui[r];++i)
    {
        if(mx[i]<mod)
            continue;
        for(int j=L[i];j<=R[i];++j)
            a[j]%=mod;
        reset(i);
    }
}

void change(int pos,int val)
{
    a[pos]=val;
    reset(dui[pos]);
}

ll qq(int l,int r)
{
    ll ans=0;
    int q1=min(R[dui[l]],r);
    for(int i=l;i<=q1;++i)
        ans+=a[i];
    if(dui[l]!=dui[r])
        for(int i=L[dui[r]];i<=r;++i)
            ans+=a[i];
    for(int i=dui[l]+1;i<dui[r];++i)
        ans+=ji[i];
    return ans;
}

int main(){

    freopen("T2.in","r",stdin);
    freopen("T2.out","w",stdout);

    scanf("%d%d",&n,&m);
    fen=(int)ceil(sqrt(n+0.5));
    for(int i=1;i<=n;++i)
    {
        scanf("%d",&a[i]);
        dui[i]=(i-1)/fen+1;
        if(mx[dui[i]]<a[i])
            mx[dui[i]]=a[i];
        ji[dui[i]]+=a[i];
    }
    for(int i=1;i<=dui[n];++i)
    {
        L[i]=(i-1)*fen+1;
        R[i]=min(n,i*fen);
    }
    int op,tin1,tin2,tin3;
    for(int i=1;i<=m;++i)
    {
        scanf("%d",&op);
        if(op==1)
        {
            scanf("%d%d",&tin1,&tin2);
            if(tin1>tin2)
                swap(tin1,tin2);
            printf("%lld\n",qq(tin1,tin2));
        }
        else
            if(op==2)
            {
                scanf("%d%d%d",&tin1,&tin2,&tin3);
                if(tin1>tin2)
                    swap(tin1,tin2);
                mm(tin1,tin2,tin3);
            }
        else
        {
            scanf("%d%d",&tin1,&tin2);
            change(tin1,tin2);
        }
    }
}

T2

 

 

 

T3

会炸long long INF 的SBT....

打表发现规律，然后 二分n即可

K=64，必须特判，不然炸了...

 

 

#include <cstdio>
#include <cstring>
#include <iostream>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;

ll num[123];

void chu(int x)
{
    mem(num,0);
    int temp=2*x;
    for(int i=x+1;i<=temp;++i)
    {
        int cc=0;
        for(int j=0;j<32;++j)
            if( i&(1<<j) )
                ++cc;
        ++num[cc];
    }
    for(int i=1;i<=32;++i)
    {
        if(num[i]==0)
            break;
        printf("%lld ",num[i]);
    }
    printf("  ");
    for(int i=7;i>=0;--i)
        printf("%d",(((1<<i)&x)?1:0) );
    printf("\n");
}

int main(){

    freopen("biao.out","w",stdout);

    for(int i=1;i<=100;++i)
        chu(i);

}

da_biao

 

 

#include <cstdio>
#include <cstring>
#include <iostream>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
const ll INF=((ll)1<<63)-1;

int T,K;
ll m;

ll C[106][106];
void chu()
{
    C[0][0]=1;
    for(int i=1;i<66;++i)
        C[i][0]=C[i][i]=1;
    for(int i=1;i<66;++i)
        for(int j=1;j<i;++j)
            C[i][j]=C[i-1][j]+C[i-1][j-1];
}

ll ji(ll x)
{
    ll ans=0;
    int cc=0;
    for(int i=63;i>=0;--i)
        if( x&((ll)1<<i) )
        {
            //printf("i=%d\n",i);
            ++cc;
            if(K-cc>=0)
                ans+=C[i][K-cc];
        }
    return ans;
}

ll erl(ll l,ll r)
{
    ll ans=INF,mid;
    ll temp;
    while(l<=r)
    {
        mid=(l+r)>>1;
        temp=ji(mid);
        if(temp==m&&ans>mid&&mid)
            ans=mid;
        if(l>=r)
            break;
        if(temp>=m)
            r=mid-1;
        else
            l=mid+1;
    }
    return ans;
}

ll err(ll l,ll r)
{
    ll ans=1,mid;
    ll temp;
    while(l<=r)
    {
        mid=(l+r)>>1;
        //printf("l=%lld r=%lld mid=%lld\n",l,r,mid);
        temp=ji(mid);
        //printf("temp=%lld\n",temp);
        if(temp==m&&ans<mid)
            ans=mid;
        if(l>=r)
            break;
        if(temp<=m)
            l=mid+1;
        else
            r=mid-1;
    }
    return ans;
}

int main(){
    
    //freopen("number10.in","r",stdin);
    //freopen("in.in","r",stdin);
    //freopen("out.out","w",stdout);

    chu();
    scanf("%d",&T);
    ll l0,r0;
    while(T--)
    {
        scanf("%lld%d",&m,&K);
        if(K==0)
        {
            printf("0 1\n");
            continue;
        }
        if(K==1)
        {
            printf("-1\n");
            continue;
        }
        if(K==64)
        {
            printf("1 %lld\n",INF);
            continue;
        }
        //K=5;
        //printf("adsd=%lld\n",ji(97));
        l0=erl(0,INF);
        r0=err(0,INF);
        printf("%lld %lld\n",l0,r0);
    }
 // cout<<(((ll)1<<63)-1);
}

T3